Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Mid-Chapter Check Point - Page 1163: 16


The value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$ is $\frac{1}{6}$.

Work Step by Step

Consider the function $ f\left( x \right)=\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$. Simplify the above function, Multiplying the numerator and the denominator of the function by $\sqrt{{{x}^{2}}+9}+3$, So, $\begin{align} & f\left( x \right)=\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}\left( \frac{\sqrt{{{x}^{2}}+9}+3}{\sqrt{{{x}^{2}}+9}+3} \right) \\ & =\frac{{{x}^{2}}+9-9}{{{x}^{2}}\left( \sqrt{{{x}^{2}}+9}+3 \right)} \\ & =\frac{{{x}^{2}}}{{{x}^{2}}\left( \sqrt{{{x}^{2}}+9}+3 \right)} \\ & =\frac{1}{\sqrt{{{x}^{2}}+9}+3} \end{align}$ The function $ g\left( x \right)={{x}^{2}}+9$ is a polynomial. Now, find the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$, $\begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\sqrt{{{x}^{2}}+9}+3} \\ & =\frac{1}{\sqrt{\underset{x\to 0}{\mathop{\lim }}\,{{x}^{2}}+9}+3} \\ & =\frac{1}{\sqrt{0+9}+3} \\ & =\frac{1}{\sqrt{9}+3} \end{align}$ $\begin{align} & =\frac{1}{3+3} \\ & =\frac{1}{6} \\ \end{align}$ Thus, the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$ is $\frac{1}{6}$.
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